#57805 Twin Paradox

[vorpal blade]

I would like to identify where we agree and disagree, as I can't always tell from what you write. Let me list some the areas where I think we agree.

1. “Where an event occurs in a single place—for example, a car crash—all observers will agree that both cars arrived at the point of impact at the same time.” http://en.wikipedia.org/wiki/Relativity_of_simultaneity

2. All clocks within an inertial reference frame can be synchronized to read the same to any observers within that frame.

3. The location of the origin for the coordinates is arbitrary and does not make any difference in the the Lorentz transformations if we use a standard configuration.

4. From the reference frame of either the outbound or the inbound spaceships the time duration of the trip for the traveling twin is 0.45 light years for each leg of the journey.

5. The earth views the traveling twin to have aged 1.5 years in the trip, and views itself as having aged 2.5 years.

If one of the twins is born and in that instant is transported to the outbound spaceship when it is closest to earth, then by the principle of simultaneity (statement 1 above) then all observers will agree to the simultaneity of this single event.

If clocks are synchronized in each reference frame, such that t=0 at the single event of transporting the twin to the spaceship, than that event will correspond to t=0 on every clock in all three reference frames. This does not mean that the incoming spaceship will see t=0 on the clock at station 2, or that someone on earth will see t=0 on the incoming spaceship when t=0 on earth. But within each reference frames the clocks will be synchronized, so that the incoming spaceship knows for a fact that the twin was born and left the earth precisely when his clock read t=0. That's a consequence of setting t=0 in his reference frame and having synchronized clocks.

If you set the origins for the coordinates on earth for each of the three reference frames at t=0, then the incoming spaceship will also see the clock on earth to read t=0 when his clock reads t=0. Setting the origin elsewhere gives him a different clock reading, indicating that clock readings are not to be taken as necessarily absolute physical facts, but relative values. However by statement 3 above it is just as correct to choose this standard configuration as any other. Since the results must be the same using any correct method, then if the theory is consistent we must interpret the clock readings in a manner to give consistent results.

Since the clock on earth appears to be at t=0 at the instant the twin is born and jumps to the spaceship, and since by constant observation of earth time shows that the clock on the earth has changed only by 0.9 years before the twin is returned, as observed by the incoming spaceship, then the traveling twin must see that the twin on earth has aged only 0.9 years.

If you calculate using the origin on the incoming spaceship you see the clock on earth appear to be at 1.6 years ahead of the clock in space station 2. This appears to those on the spaceship as a lack of synchronization of clocks. You can still see the elapsed time on earth as being the difference between 2.5 years, when you pass the earth, and 1.6 when the twin left the earth in this particular choice of coordinate origin. The twin cannot be 1.6 years old when leaving the earth, as the simultaneous event of twin being born and jumping to the spaceship is observed by all as a simultaneous event.

I believe I am taking into account all of the steps of relativity, not leaving anything out, and not choosing special conditions which give me the result I want. If you don't think so then let me know specifically what I am leaving out.

Let me know which of these statements you disagree with, why. I'll assume you agree unless stated otherwise.

[Tao]

Awesome. If we work with t=0 based off of that and not off of the x coordinate system, things might be much clearer. Forgive me for misconstruing your intent and making things more complicated.

mmkay. So we're in agreement over the view from earth. And the view from the outbound spaceship is pretty set as well, I think. For the return spaceship: at t=0 it will view the outbound passing earth and receiving it's passenger 3.33 ly away. The earth is moving towards it at .8c and the outbound spaceship at .97560c. {relativistic velocity addition is v+u/(1+(vu/c²)), so .8c of earth and the .8c of outbound = .97560c to the return ship}

3.4166 years later, (3.33/.97560= 3.4166) the outbound spaceship meets it, carrying an infant that has only seen .75 years (lambda of .97560c is 4.55, Taking 3.4166/4.55 gives us .75)

At the 4.1666 year mark (3.33/.8=4.1666) Earth arrives. The infant has aged yet another .75 years (4.1666-3.41666=.75) Earth has aged 2.5 years in the process. (as we've seen lambda of .8c is 1.66 and 4.1666/1.666 is 2.5 years)

So, we've proved it yet again.

The only thing left to clear up that I can foresee is the fact that the returning spaceship is at 3.33ly away at its t=0.

From earth it appears to be 2 ly away and approaching at .8c. Shouldn't this length dilate down to 1.2 ly? Well, if you're working off of earth's time, yeah, you can (and actually have to) do that. But from it's own perspective when it is 1.2 ly away from earth, the shuttle has already left earth long ago and is quickly approaching. By our defining of t=0 being that transition in that reference frame, we can definitely say that at its t=0 it cannot be at 1.2ly away.

One quick way to prove this would to look at it from the outbound ship's point of view. From the stationary outbound ship's point of view at t=0 earth is immediately under them and the return ship is it is 0.731ly away and approaching at .97560c. (.97560c * .75ly is .731ly)

Mmm, more to come I'm sure, but I've got to go right now. Needless to say, freeing the time coordinate from the space coordinate sure makes the math easier.

[vorpal blade]

Well, you didn’t answer any of my questions, or respond to my comments, but your new scenario is interesting. I like it better than your old scenarios which weren’t correct when you tried to interpret clock readings as elapsed time and added clock readings to get elapsed time, or tacked on an additional 1.6 years at the start for the twin who stayed on earth, or added an additional 1.6 years at the end for the sake of simultaneity. I can see why you stopped using Lorentz coordinate transforms. This scenario is cleverer in that you add the 1.6 years gradually to the twin on earth by making the return ship travel a very long distance before coming to the scene of action.

In the new scenario the twin on earth is seen by those on the return ship to age 3.4166/1.666 = 2.05 years before the outbound spaceship meets the incoming spaceship and the traveling twin is transferred. The problem with your new scenario is that we don’t really care what those on the incoming spaceship have seen regarding the aging of the stay-at-home twin. What matters is what the traveling twin sees. And what that twin has seen is the twin on earth age 0.45 years while the traveling twin is outbound. On the return trip he sees the twin on earth age (4.1666-3.41666=.75)/1.666 = 0.45 years. Thus the traveling twin sees the stay-at-home twin age a total of 0.45+0.45 = 0.9 years while the traveling twin is traveling.

We’ve proved it yet again, each twin sees the other twin age less than himself. The twin paradox remains.

[Tao]

Well, you didn’t answer any of my questions, or respond to my comments, but your new scenario is interesting.

This was in direct response to my realization that your setting of the origin was only applicable to space co-ordinates, not time. If your timeline is fixed at a arbitrary point, the Lorentz transformations end up looking just as I did them, and the results are the same, if convoluted to get to. That convolution is part of the reason I argued against such a fixed perspective when it was first brought up.

I like it better than your old scenarios which weren’t correct when you tried to interpret clock readings as elapsed time and added clock readings to get elapsed time

as I've linked before, the equations that are easiest to post deal with time intervals, with beginnings already subtracted from ends, just as I dealt with lengths in light years not in "point a less point b". If you are still getting hung up on that, we can run the same numbers through T=t₂-t₁=(t'₂+(vx'₂/c²)-t₁+(vx'₁/c²))/((1-(v²/c²))^1/2). Which, since we are allowing time measurements to be made from stationary locations, works out to T=T₀/((1-(v²/c²))^1/2)=T₀γ, which is what I've been doing. (Note that T=t₂-t₁is a interval, taking reading subtraction into account already, forgive me for not showing each step of my work, my professors would get after me for the same thing. I just assume that things that make the numbers easier to crunch and easier to grade would be preferred, as long as I don't screw up anything in the assumed step, a difficult task when something as easy as "T by gamma" (I called it lambda in my previous post, once again showing my screwups, γ is gamma λ is lambda, a rookie mistake considering my interest in the language.))

I can see why you stopped using Lorentz coordinate transforms. This scenario is cleverer in that you add the 1.6 years gradually to the twin on earth by making the return ship travel a very long distance before coming to the scene of action.

The Lorentz transforms are still there, you cannot work relativistically without them. As for the length, I've explained before why it is so far out. By your own declaration, I was able to get away from my misunderstanding that I was forced to set my timeline at zero when earth was 2 mi from the ship in it's view and 1.2 from the ship in the ships view. Now that I understand you would like to set the timeline at zero when the outbound ship passes earth, I did so. There is no manipulation on my part to make it work out that way, that is where it would be at t=0. If you'd refer a couple posts back, you gave me a good example of how to show this. If you made the return ship exceedingly long, with the extreme tail being the portion carrying the baby that we've considered as the ship up till now, how long would it have to be at rest in order for the nose to be passing earth at the same time as the outbound ship? However long it is at rest would be how far earth is in the reference frame of the return ship, at the moment of departure. If it is anything other than 1.2 ly long, we would have to do something to make up that discrepancy when looking at setting t=0 to that location as opposed to 1.2 ly. Perhaps that would explain the 1.6 ly of the Lorentz shift better than anything I've tried thus far. As you've already mentioned, we can set the origin of the time line anywhere you want and we could run the equations and get the same answers.

In the new scenario the twin on earth is seen by those on the return ship to age 3.4166/1.666 = 2.05 years before the outbound spaceship meets the incoming spaceship and the traveling twin is transferred. The problem with your new scenario is that we don’t really care what those on the incoming spaceship have seen regarding the aging of the stay-at-home twin. What matters is what the traveling twin sees. And what that twin has seen is the twin on earth age 0.45 years while the traveling twin is outbound. On the return trip he sees the twin on earth age (4.1666-3.41666=.75)/1.666 = 0.45 years. Thus the traveling twin sees the stay-at-home twin age a total of 0.45+0.45 = 0.9 years while the traveling twin is traveling.

So what you are wanting to do is reduce the earth's twins' age by a sudden jump when the return ship takes on it's passenger? What would the pilot of such a ship see? When they arrive, he perceives a child halfway through it's third year but his infant passenger would see a child but a few months old? This isn't paradoxical to you? Would the ship and child be out of phase with each other like a Star Trek episode? Yes, you can claim you don't care to take note of portions of a mathematical equation, and you are perfectly fine in doing so, but your choice of not doing the math does not reflect on the consistency of the process.

The child ages at it's own rate, as if it were stationary at all times, but its timeline relative to the rest of the universe is based upon it's reference frame, you cannot ignore it just because you'd rather there be a flaw in the theory. Does this mean that in its perspective, the earth child ages rapidly as it changes ships? Yes. Yes it does. This is the lack of symmetry that others point to showing how the paradox isn't one. The earth child stays in one reference frame, the traveler doesn't. Much of the apparent aging occurs when changing reference frames. In a GR stance, this occurs during deceleration and again at acceleration back towards the Earth. Ignoring these effects can be construed to form a paradox, but so can ignoring any law of mathematics (If we ignore the divide by zero rule I can prove that 2+2=5! Addition isn't internally consistent!) We didn't want to deal with the calculus of GR, (and I'm not sure I'd be able to get LaTeX equations up on this forum, more my problem than that of the forum), but since everything can be proven w/o GR, we're fine. (And, as I've said, the problem in GR has been broken down to it's general form, showing for all variables, the non accelerated twin is younger).

Forgive me for not answering your previous post in a step by step manner, it was my birthday and my parents were kind enough to take me out to eat. That took precedence at that moment. I am similarly pressed for time now, as I'm driving illegally until I can get to the courthouse and renew my drivers' license. I am in no way trying to ignore you, and apologize if that is the way I come across.

[Tao]

MMkay, now I've some time to get to your post.

I would like to identify where we agree and disagree, as I can't always tell from what you write. Let me list some the areas where I think we agree.

1. “Where an event occurs in a single place—for example, a car crash—all observers will agree that both cars arrived at the point of impact at the same time.” http://en.wikipedia.org/wiki/Relativity_of_simultaneity

Agreed.

2. All clocks within an inertial reference frame can be synchronized to read the same to any observers within that frame.

Agreed.

3. The location of the origin for the coordinates is arbitrary and does not make any difference in the the Lorentz transformations if we use a standard configuration.

Agreed. Though placement of x or t coordinates does affect the work done, the results end up the same, as we've seen through my misunderstanding.

4. From the reference frame of either the outbound or the inbound spaceships the time duration of the trip for the traveling twin is 0.45 light years for each leg of the journey.

Close. They'd see each leg while carrying the child as 0.75 years in duration. if you were looking for the number of light years the Earth travels while each ship is carrying the child, it would be 0.6 ly each way.

5. The earth views the traveling twin to have aged 1.5 years in the trip, and views itself as having aged 2.5 years.

Agreed again.

If one of the twins is born and in that instant is transported to the outbound spaceship when it is closest to earth, then by the principle of simultaneity (statement 1 above) then all observers will agree to the simultaneity of this single event.

Agreed.

If clocks are synchronized in each reference frame, such that t=0 at the single event of transporting the twin to the spaceship, than that event will correspond to t=0 on every clock in all three reference frames. This does not mean that the incoming spaceship will see t=0 on the clock at station 2, or that someone on earth will see t=0 on the incoming spaceship when t=0 on earth. But within each reference frames the clocks will be synchronized, so that the incoming spaceship knows for a fact that the twin was born and left the earth precisely when his clock read t=0. That's a consequence of setting t=0 in his reference frame and having synchronized clocks.

Agreed. It also makes the math easier.

If you set the origins for the coordinates on earth for each of the three reference frames at t=0, then the incoming spaceship will also see the clock on earth to read t=0 when his clock reads t=0. Setting the origin elsewhere gives him a different clock reading, indicating that clock readings are not to be taken as necessarily absolute physical facts, but relative values. However by statement 3 above it is just as correct to choose this standard configuration as any other. Since the results must be the same using any correct method, then if the theory is consistent we must interpret the clock readings in a manner to give consistent results.

Agreed.

Since the clock on earth appears to be at t=0 at the instant the twin is born and jumps to the spaceship, and since by constant observation of earth time shows that the clock on the earth has changed only by 0.9 years before the twin is returned, as observed by the incoming spaceship, then the traveling twin must see that the twin on earth has aged only 0.9 years.

Nope. By changing inertial reference frames, The infant's perception of the earth child's age changes. While only .45 at the moment transfering off of the receding ship, the effects of changing inertial reference frames (in addition to squishing him flat, but we can assume he is very resilient) is to observe that which things in his new inertial reference frame also observe. To declare otherwise would be illogical, but thankfully, the math backs us up on this. At the moment the child begins returning to the earth, it would see its twin as 2.05 years old.

If you calculate using the origin on the incoming spaceship you see the clock on earth appear to be at 1.6 years ahead of the clock in space station 2. This appears to those on the spaceship as a lack of synchronization of clocks. You can still see the elapsed time on earth as being the difference between 2.5 years, when you pass the earth, and 1.6 when the twin left the earth in this particular choice of coordinate origin. The twin cannot be 1.6 years old when leaving the earth, as the simultaneous event of twin being born and jumping to the spaceship is observed by all as a simultaneous event.

Agreed, somewhat. The clock on earth would be reading 1.6 when Space Station two passes the return ship. But the return ship would not see the outbound ship passing earth as space station two passes it. When space station two passes it, it would read 2.6666 on it's own clocks, Earth would be 1.2 light years away reading 1.6y and the oncoming ship would be 0.468ly from the earth, carrying a baby of 0.5853 years.

Note that the time and distance of Earth matches what we got when I understand Space station two's passage as being the origin of the t coordinate.

Now for the math there. We can calculate the contracted distance between space station two and the return ship, or we can work with the knowledge that it takes .75 years between space station one's passage and earth's and the distance between them is identical. That gives us 1.5 years of travel time before the earth passes. Since we know the earth passes at the 4.1666 year mark (3.33/.8=4.1666) we can guess SS2 will pass at the 2.66666 year mark. Running a fact check by doing it the straightforward way gives us the earth 3.333ly away at t=0 and SS2 at 2.1333 ly away (the 2ly from earth and SS2's rest frame gets contracted to 1.2 miles, since they're moving at .8c, {please don't ask me to work that out, I think we've established that by now} 3.333-1.2=2.133) 2.1333 ly at .8c is 2.6666 years. Good. That's established. So it took 2.666 years of the returning ship's time to get to SS2.

How much time has elapsed on earth? 2.666 divided by 1.66666 (1.666 is one over root(1-(v²/c²)) or the gamma of .8c,) is 1.6years. How far has the earth traveled? .8c by 2.666years is 2.133, less the starting point of 3.333 is 1.2 ly.

How much time has elapsed on the outbound spaceship? 2.666 divided by 4.555 is .5853 years. (Once again 4.555 is the gamma of .97560c and .97560c is the speed of the oncoming outbound spaceship relative to the stationary returning one) The traveling babe is .5853 years old. How far has the ship traveled? .97560 by 2.666 years is 2.6016ly. That from the starting point of 3.333 is .7317 ly away from the ship and .468 ly from the earth.

I believe I am taking into account all of the steps of relativity, not leaving anything out, and not choosing special conditions which give me the result I want. If you don't think so then let me know specifically what I am leaving out.

Let me know which of these statements you disagree with, why. I'll assume you agree unless stated otherwise.

Posting now as it's blowing hard and I'd rather not lose power/internet and be rendered unable to.

[Tao]

You've called this a new scenario, but I'd like to point out that it isn't at all. Previously, I had understood you're desire to set the origin at earth's location as being unreasonable, but doable. This is due in no small part to the way I crunch the numbers. Since you had asked for t=0 to be the same in all reference points, and the origin to be set at earth, I did my best to stick to it, even though it made little sense to me. Earth's distance from the return ship from it's point of view when the babes are born (When t=0 in the earth's perspective) was 1.2 ly from the return ship and 0 ly from the outbound ship, so I set that as my origin, and set t=0 when the earth passes that point in any reference frame.

I'll try to set it up graphically so you can see what was going through my head.

Key:
> Outbound spaceship
< Return spaceship
E Earth
| space stations
. markers to make things easier, of no consequence
o origin.


-------------------------------------------<
--------------------->----------------------
----------.----------E----------|----------|
----------.----------o----------1ly
Beginning

--------------------------------<-----------
-------------------------------->-----------
----------.----------E----------|----------|
----------.----------o
Transfer

---------------------<----------------------
------------------------------------------->
----------.----------E----------|----------|
----------.----------o
Finish




The outbound spaceship wouldn't see things quite the same, but not much differently.

------------------------------------<-------------------
--------------------->---------------------------------
----------.----------E----------|----------|----------|
----------.----------o----------.6ly
Beginning (note oncoming ship's location, I think it should be .73ly away, at .9756c this puts transfer at .6 ly away from earth in .75 years, which seems to mesh.)

---------------------<---------------------------------
--------------------->---------------------------------
----------E----------|----------|----------|----------|
--------.6ly----------o
Transfer (note that the origin fails to move as the earth races away. I had assumed by demanding that the origin be set at earth, you wanted it at x=0 t=0 as the earth perceived it. Since this happens to be co-located with the outbound spaceship, it has little to no impact on these equations.)

---<----------------------------------------------------
--------------------->---------------------------------
---E----------|----------|----------|----------|-------
1.05----------.------o
Finish. Notice that this is not .75 years later, but .615, earth has only gone another 0.492 ly for a total of 1.092




Now for the fun one...

-------------------------------------------<-----------
----------------------------->------------------------
----------.----------E----------|----------|----------|
----------.----------o----------.6ly
Beginning, with the earth at the origin, the return ship at 1.2ly away and the outbound ship at .468 ly from the earth. Now, on this diagram, Earth had already seen 1.6 ly since the outbound ship took off, according to the returning ship. Yet since I had been under the impression that this is what you wanted: the origin at earth, and this as t=0 for the return ship, I had do do some work to make any sense of it. To do so, I had to run a Lorentz transformation on the distance between the earth and the returning ship, (a seemingly arbitrary distance of 1.2ly, but hey, you can set the origin at any arbitrary distance and still have the numbers gel.) so, with an x value of 1.2 and a speed of .8c I was getting t=0 for the ship is 1.6 years for earth (which is what earth's clock is reading here)

-------------------------------------------<-----------
------------------------------------------->----------
----------.----------.----------E----------|----------|
----------.----------o----------.6ly
Transfer. 0.75 years later, the oncoming ship, traveling at .97560c has traveled the .7317ly and met up, the earth is now .6 ly away and approaching at .8c. (Note that this leg does indeed take .45years earth time, but since the outbound ship wasn't at the earth at the beginning, the earth infant's age must be older than .45years.)

-------------------------------------------<-----------
--------------------------------------------------------
----------.----------.----------.----------E----------|
----------.----------o----------.----------1.2ly
Finish. The origin is still 1.2 ly away, the earth has reached the return ship, the outbound ship is well past the second space shuttle. Another .75 years have gone by for the traveler, and another .45 for the babe on earth. 0.45 here plus the 0.45 on the previous step, and the 1.6years needed to get to the beginning step from birth gives us the age 2.5 that all parties would agree on.




Hopefully that clears up something. Notice that the origin does not move and is at earth at t=0 (beginning) of each run.

This is Non-standard, as you've stated before. It definitely is not the best way to do this, and is why I argued against setting t=0 and x=0 at earth in all perspectives. But you have to admit, the math does work out, even with such an arbitrary (again, my fault, not yours) setup.

You can see from the 'Beginning' in the return ship's point of view that it was some ways off when > passed E. By your previous post freeing t=0 from x=0 in my mind, it is much easier to start and run the math for each step, but the situation is the exact same.

[Tao]

Holy wall of text, batman. That's over 2k new(non-quoted) words added today, bringing just my total up over 11k. Between the two of us, we've dang near written a novel!

[vorpal blade]

I’m sorry, I gave you a hard time in my last post. I apologize.

Shortly after my last post I discovered that my calculations posted on June 15th are in error. The problem I first discovered is in the calculation for what the clock should read in the outgoing spaceship at the time the twin is transferred from the outgoing to incoming spaceship. The problem is that I had the right answer, but the values I inputted into the equation are wrong. Had I noticed this earlier I would have concluded that something was fundamentally wrong with my calculation of the position of the return ship at t=0. But the misleading “correct answer” obscured the problem from my view.

Unfortunately neither of us noticed this error. And instead of pointing out that what was intended as my calculation for the starting position of the return ship was inconsistent with my assumptions, you merely changed all my starting assumptions (the outgoing spaceship is at the earth at the start, the clocks read zero at the start within each reference frame, the twins are born at the start) to be consistent with my erroneously calculated value for the starting position of the return ship. I did not know this at the time, and it looked like to me you were refusing to understand or use my starting assumptions, which I considered the standard configuration.

So, I no longer have any problem with how you did the calculation. I no longer believe I have an alternate correct approach that gets a different answer. Changing the origin does not seem to make a difference, as long as you consider clocks as giving relative values.

Let me clear up a few things, if I can. The first doesn’t make any difference, since I am not disagreeing with you, but is for the record. Previously (July 23) I posted 5 points that I thought we would agree upon. I can see that the 4th point doesn’t say what I meant to say. What I meant was that “From the reference frame of either the outbound or the inbound spaceships the time duration of the trip for the stay-at-home twin is 0.45 years for each leg of the journey.” I believe you have already agreed to that.

I do disagree with you on the time it takes for the incoming spaceship to reach earth as seen from the outgoing spaceship. Here is what you said:

---<----------------------------------------------------
--------------------->---------------------------------
---E----------|----------|----------|----------|-------
1.05----------.------o
Finish. Notice that this is not .75 years later, but .615, earth has only gone another 0.492 ly for a total of 1.092

I believe that the correct value is not 0.75 years later, or even 0.615 years later, but 3.4166 years later. I calculate this on the basis that the incoming spaceship appears to be passing the outgoing spaceship at the velocity 0.9756c headed toward earth, which is 0.6 light years away as seen by the outgoing spaceship. However, the earth appears to be moving away from the outgoing spaceship at the velocity 0.8c. Therefore the incoming spaceship appears to be gaining on the earth as the difference of velocities, 0.9756c – 0.8c = 0.1756c. At that rate it will take 0.6/0.1756 = 3.4166 years to catch up to earth. This is not too surprising, as you pointed out that in the first 0.75 years the incoming spaceship got only 0.73-0.6 = 0.13 light years closer to earth.

An additional interesting fact is that in the first 0.75 years, as seen on the outgoing spaceship, those on earth appear to have aged only 0.45 years. The twin leaves this spaceship and then for the next 3.4166 years those on the outgoing spaceship watch the twin try to catch up to earth. During that time they see the earth age an additional 3.4166/1.666 = 2.05 years. You have to admit that the problem has symmetry.

I checked all your other calculations in yesterday’s posts, and it looks to me that you did them correctly. I think some confusion still exists, but hopefully it is not important.

[vorpal blade]

Holy wall of text, batman. That's over 2k new(non-quoted) words added today, bringing just my total up over 11k. Between the two of us, we've dang near written a novel!

The bad thing is that writing so much discourages others from joining into the conversation. Who has the many hours it would take to read and digest so much, just to come up to speed and feel like they are following along? I try to be brief, and I think you do to, but sometimes writing less only causes more confusion.

[vorpal blade]

So what you are wanting to do is reduce the earth's twins' age by a sudden jump when the return ship takes on it's passenger? What would the pilot of such a ship see? When they arrive, he perceives a child halfway through it's third year but his infant passenger would see a child but a few months old? This isn't paradoxical to you? Would the ship and child be out of phase with each other like a Star Trek episode? Yes, you can claim you don't care to take note of portions of a mathematical equation, and you are perfectly fine in doing so, but your choice of not doing the math does not reflect on the consistency of the process.

The child ages at it's own rate, as if it were stationary at all times, but its timeline relative to the rest of the universe is based upon it's reference frame, you cannot ignore it just because you'd rather there be a flaw in the theory. Does this mean that in its perspective, the earth child ages rapidly as it changes ships? Yes. Yes it does. This is the lack of symmetry that others point to showing how the paradox isn't one. The earth child stays in one reference frame, the traveler doesn't. Much of the apparent aging occurs when changing reference frames. In a GR stance, this occurs during deceleration and again at acceleration back towards the Earth. Ignoring these effects can be construed to form a paradox, but so can ignoring any law of mathematics (If we ignore the divide by zero rule I can prove that 2+2=5! Addition isn't internally consistent!) We didn't want to deal with the calculus of GR, (and I'm not sure I'd be able to get LaTeX equations up on this forum, more my problem than that of the forum), but since everything can be proven w/o GR, we're fine. (And, as I've said, the problem in GR has been broken down to it's general form, showing for all variables, the non accelerated twin is younger).

Hmm, is that what you think, that I merely claim to not care to take note of portions of a mathematical equation? That I ignore the facts because I’d rather believe the theory has a flaw? That I am ignoring the legitimate effects of acceleration and deceleration in order to form a paradox? You seem to have a pretty poor opinion of me. Well, we know lots of others who share your opinion of me. You are nicer than most in expressing it, however.

I doubt that I can change your opinion of me, but when I have some more time I’d like to say a few things for the record about what I’m thinking in regard to this change in perception as one jumps reference frames. And I’d like to hear your opinion about the meaning of what is happening. I’ll ask some specific questions later. As always, respond when you have time, and if you wish. There is no rush, except I could quit this forum at any time, and considering my generally low approval level that is a real possibility.

Thank you for your time.

[Tao]

I do disagree with you on the time it takes for the incoming spaceship to reach earth as seen from the outgoing spaceship. Here is what you said:

---<----------------------------------------------------
--------------------->---------------------------------
---E----------|----------|----------|----------|-------
1.05----------.------o
Finish. Notice that this is not .75 years later, but .615, earth has only gone another 0.492 ly for a total of 1.092

I believe that the correct value is not 0.75 years later, or even 0.615 years later, but 3.4166 years later. I calculate this on the basis that the incoming spaceship appears to be passing the outgoing spaceship at the velocity 0.9756c headed toward earth, which is 0.6 light years away as seen by the outgoing spaceship. However, the earth appears to be moving away from the outgoing spaceship at the velocity 0.8c. Therefore the incoming spaceship appears to be gaining on the earth as the difference of velocities, 0.9756c – 0.8c = 0.1756c. At that rate it will take 0.6/0.1756 = 3.4166 years to catch up to earth. This is not too surprising, as you pointed out that in the first 0.75 years the incoming spaceship got only 0.73-0.6 = 0.13 light years closer to earth.

This wouldn't surprise me, upon seeing my numbers singled out they don't look right. I've not the time tonight to crunch them and see what if anything went wrong.

An additional interesting fact is that in the first 0.75 years, as seen on the outgoing spaceship, those on earth appear to have aged only 0.45 years. The twin leaves this spaceship and then for the next 3.4166 years those on the outgoing spaceship watch the twin try to catch up to earth. During that time they see the earth age an additional 3.4166/1.666 = 2.05 years. You have to admit that the problem has symmetry.

Again, w/o checking your numbers, I'd say they feel right. The lack of symmetry is really only when comparing the experiences of the two children, as the one experiences a shift in inertial reference frames, while the other doesn't.

Who has the many hours it would take to read and digest so much, just to come up to speed and feel like they are following along? I try to be brief, and I think you do to, but sometimes writing less only causes more confusion.

Too true.

Hmm, is that what you think, that I merely claim to not care to take note of portions of a mathematical equation? That I ignore the facts because I’d rather believe the theory has a flaw? That I am ignoring the legitimate effects of acceleration and deceleration in order to form a paradox? You seem to have a pretty poor opinion of me.

Heh, not usually. I'll admit, there are times when it seems like that would be an easier explanation than admitting my own inability to explain what seems simple to me. Remember I'm largely self-taught. To see a statement of "We don't care about X" or "Y shouldn't be a factor" when they are crucial steps in my understanding of something should merely be flagging a topic where we disagree, but it usually requires me to shift my entire mindset of what is plainly given and what is more enigmatic.

[vorpal blade]

I can see now how the twin paradox can be explained using the Special Theory of Relativity, in particular using the Lorentz transformations. I had thought that it couldn’t be done without the General Theory, and doubtful even then, but I was mistaken. Thank you very much for helping me to see this.

Everyone is agreed that the key to resolving the paradox is to note that only one twin is accelerated. The problem I had was that just because one twin feels acceleration does not automatically mean that he must be the one to age more slowly, or that we know a priori that acceleration makes any difference at all. If you say that we can solve this problem without involving acceleration it sounds like acceleration can be neglected. But if acceleration can be neglected that we can imagine that the stay-at-home twin was accelerated and should therefore age less than the other twin. I think the answer is that you can’t neglect acceleration, because acceleration is just a rate of change of velocity, and we know that velocity affects time, according to Einstein’s theory. So, acceleration is really important, if for no other reason than to determine which way to go in making an absolute difference come from the principle of relativity.

One of the conceptual problems I have with the Lorentz coordinate transformations is the problem of a predetermined future. For someone in the universe, travelling at great speed toward the earth, but at a long distance away, our future is already past history. I think it is astounding that such a profound philosophical and theological conclusion should come from a straightforward application of the principle of relativity and the speed of light being a constant.

I think it is one thing to picture this as just a “perception,” and not reality, but evidently if the twin-absolute-time difference is real then this is not jut perception. You gave the analogy of two people at a distance who see each other grow smaller as the distance increases. Yet we know that if we bring them back together they will be the same size again. The twin paradox tells us that one (and only one) of the two people moving away from us will actually be younger, or in terms of the analogy, will actually be smaller when reunited with us. To think that movement should make such a difference on absolute time!

In our example I was curious to know whether the returning spaceship could actually see into the future. Perhaps someone from this spaceship could jump to the space station, or the outgoing spaceship and tell them something about the future. It turns out, if I did the calculations correctly, that this is not the case. Since the speed of light is assumed to be a constant the returning spaceship can’t actually see the spaceship fly by earth and pick up the twin until about 1/12 year before arriving at the space station and receiving the traveling twin. At the instant the returning spaceship passes the space station the latest videos (or telescopic images) would show the twin on earth to be 0.25 years old. That is exactly the same picture those on the space station would see, and those on the outgoing spaceship. So, if you judged by what the telescopes (or video images transmitted at the speed of light) show, the twin on the outgoing spaceship sees the twin on earth age from 0 to 0.25 years during the 0.75 years he is on that spaceship. He transfers to the incoming spaceship and the aging goes on smoothly from 0.25 years to 2.5 years when the returning spaceship comes to earth.

One of the curious things about this is that those on the incoming spaceship only calculate that the twin on the earth is aging more slowly. Looking at the video which I imagine is streaming live from earth, they actually see the twin age three times faster on earth than in the spaceship. In 0.75 years, the time it takes to go from space station 1 to earth, the images from earth are Doppler shifted to show aging from 0.25 years to 2.5 years. That speeded up apparent aging is in contrast to the calculated slow aging from 2.05 years (actually ahead of the earth twin in proper time) to 2.5 years. I find that fascinating.

I’ve been trying to understand the calculations using acceleration and General Relativity that you referenced. So far I haven’t been able to understand it sufficiently to see how the equations were derived. But I’m working on it. The equations for the twin who travels, as seen from earth, is easy to calculate and easy to believe. I wish the equations from the point of view of the traveling twin were given in the same manner, phase by phase. But the equations representing what the traveling twin sees are much harder to interpret and understand.

If we are to assume that the Lorentz equations that we have used accurately depict what happens, and that the General Theory just adds a little refinement or detail, then the reality is that for the twin who travels the twin on earth ages more slowly during the times of constant velocity. The discrepancy in time, to make the traveling twin age more slowly, is more than made up when the traveling twin jumps reference frames. I’m having some trouble seeing how the given equations using General Relativity show that the stay-at-home twin ages more slowly while the traveling twin is in constant velocity.

So, I wonder if you could comment on: (1) how it is possible that our future is already past history for others in the universe? (2) If the observations are that someone is aging faster in a moving reference frame, how do we know they are really aging more slowly? [Actually, I think I can answer that question, but I’m curious how you answer it.] And (3) how do the calculations (in the refernced article) using General Relativity show that the stay-at-home twin ages more slowly during times of constant velocity? Also, (4) would you be interested in discussing criticisms of the experiments supposedly proving Special Relativity?